The reaction 2NO + Br(2) to 2NOBr, obeys...

The reaction `2NO + Br_(2) to 2NOBr`, obeys the following mechanism:
`NO + Br_(2) overset("fast")(hArr) NOBr_(2), NOBr_(2) + NO overset("Slow")(to) 2NOBr` The rate expression of the above reaction can be written as

`r = k [NO]^(2) [Br_(2)]_(2)`

`r = k [NO][Br_(2)]`

`r = k [NOBr_(2)]`

Verified by Experts

The correct Answer is:

For slowest step : rate `= k [NOBr_(2)][NO]` …(i)
For equilibrium also `K_(c ) = ([NOBr_(2)])/([NO][Nr_(2)])`
Rate `= k [NO]^(2) [Br_(2)]`

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The rate and mechamical reaction are studied in chemical kinetics. The elementary reactions are single step reaction having no mechanism. The order of reaction and molecularity are same for elementary reactions. The rate of forward reaction aA + bBrarr cC+dD is given as: rate =((dx)/(dt))=-1/a(d[A])/(dt)=-1/b(d[B])/(dt)=1/c(d[C])/(dt)=1/d(d[D])/(dt) or expression can be written as : rate =K_(1)[A]^(a)[B]^(b)-K_(2)[C]^(c )[D]^(d) . At equilibrium, rate = 0 . The constants K, K_(1), K_(2) are rate constants of respective reaction. In case of reactions governed by two or more steps reaction mechanism, the rate is given by the slowest step of mechanism. For the reaction, 2NO+Br_(2) hArr 2NOBr , the mechanism is given in two steps: (1) NO+Br_(2) overset("fast")(hArr)NOBr_(2) (2) NOBr_(2)+NO overset("slow")(rarr)2NOBr The rate expression for the reaction is:

(a) `r=K[NO]^(2)[Br_(2)]`

(b) `r=K[NO][Br_(2)]`

(c ) `r=K[NO][Br_(2)]^(2)`

(d) `r=K[NOBr_(2)]`

`r=K[NO_(2)]^(2)[F_(2)]`

`r=K[NO_(2)]`

`r=K[NO_(2)][F_(2)]`

`r=K[F_(2)]`

The order of the reaction with respect to NO is two :

The molecularity of the steps (1) and (2) are two each

The molecularity of the overall reaction is three.

The overall order of the reaction is three